We are given the nonlinear system:
$\tag 1 x' = y, ~~ y'= -8 \sin x - 2y,$
where $-2\pi \le x \le 2\pi$.
We start off by finding the fixed points of the system. To do this, we need to find those points where $x'$ and $y'$ are simultaneously equal to zero, over the given range in $(1)$.
So, $x' = 0$, when $y = 0$, and $y' = -8 \sin x - 2y = 0$ reduces to (since we must have $y = 0$ from the previous observation):
$y' = -8 \sin x -2(0) = 0 \rightarrow x = -2\pi, -\pi, 0, \pi, 2\pi$.
This gives us a total of five fixed (critical) points as:
$$(-2\pi, 0), (-\pi, 0), (0,0), (\pi, 0), (2\pi, 0).$$
In order to linearize the system, we have to investigate the behavior of the Jacobian of the system at those fixed points. So, the Jacobian matrix is:
$$A = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ - 8 \cos x & -2 \end{bmatrix}$$
Next, we need to evaluate the Jacobian matrix $A$ at each of those fixed points.
At $(-2\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (-2\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$
At $(-\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (-\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$$
At $(0, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (0) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$
At $(\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$$
At $(2\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (2\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$
Notice, we have two identical sets of matrices given the periodicity of $\cos x$ over our five fixed points.
We are asked to linearize the system near every equilibrium point, and describe the behaviour of the linearized system. So, now we need to determine the behavior of these two matrices by looking at their eigenvalues. So, we get:
$A = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$, has eigenvalues $\lambda_1 = -4$ and $\lambda_2 = 2$. These are real eigenvalues with opposite sign, so this is an unstable saddle node.
$A = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$, has eigenvalues $\lambda_1 = -1 + \sqrt{7}i$ and $\lambda_2 = -1 - \sqrt{7}i$. These are a complex conjugate pair, with negative real part, so this is a stable spiral point.
Note: The only thing that allows us to use this linearization is that we do not get borderline cases from the fixed points. You had better make sure you are clear on this last statement!
Lastly, we can draw the phase portrait to show the direction field, the five fixed points and many solutions $x(t)$ and $y(t)$ in order to visualize this and to compare to our analyses. Of course, we should see two unstable saddle nodes and three stable spirals from the analyses above.
Make sure to pick out the five fixed points $(x, y)$ and that you see what we derived.
Regards
The basic process is to find the critical points, evaluate each critical point by finding eigenvalues/eigenvectors using the Jacobian, determine and plot $x$ and $y$ nullclines, plot some direction fields and use all of this type of information to draw the phase portrait.
You can see two different views of this process at this website and notes.
For your particular problem
$$x' = 2 - 8x^2-2y^2 \\ y' = 6xy$$
We find the critical points where we simultaneously get $x' = 0, y' = 0$ so
$$(x, y) = (0, -1), (0, 1), \left(-\dfrac{1}{2}, 0\right), \left(\dfrac{1}{2}, 0\right)$$
The Jacobian is
$$J(x, y) = \begin{bmatrix}\dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y}\\\dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y}\end{bmatrix} = \begin{bmatrix}-16 x & -4y\\6y & 6x\end{bmatrix}$$
Evaluate eigenvalue/eigenvector for each critical point
$J(0, -1) \implies \lambda_{1,2} = \pm 2 i \sqrt{6}, v_{1,2} = \left(\mp i \sqrt{\frac{2}{3}}, 1\right) \implies$ spiral
$J(0, 1) \implies \lambda_{1,2} = \pm 2 i \sqrt{6}, v_{1,2} = \left(\pm i \sqrt{\frac{2}{3}}, 1\right) \implies$ spiral
$J(-\frac{1}{2}, 0) \implies \lambda_{1,2} = (8, -3), v_{1} = (1,0), v_2 = (0, 1) \implies$ saddle
$J(\frac{1}{2}, 0) \implies \lambda_{1,2} = (-8, 3), v_{1} = (1,0), v_2 = (0, 1) \implies$ saddle
Using all the above (critical points, eigenvalues/eigenvectors, x-nullcline (red and black curves), y-nullcline (green curve), direction fields, etc.), you can now sketch the phase portrait. Exercise - make sure to add direction fields from the two sets of notes linked above so you understand how to do that. The phase portrait will look like:
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